Periodic thresholds for divisor windows

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For Problem 450, fix an integer $n\ge2$ and require a bound for every starting point and every sufficiently large length. Put $S=\{n+1,\ldots,2n-1\}$, $M=\operatorname{lcm}(S)$, and let $b(m)$ indicate that some $d\in S$ divides the integer $m$. If $C$ residues modulo $M$ are marked, their density is $\delta=C/M$, with $0<\delta<1$. The following are complete author-recorded deductions by Plasma AI, not a sharp asymptotic estimate for the threshold.

For real $x$ and $y>0$, splitting $(x,x+y)$ into periods and a remainder bounds its marked count by $\delta y+M$. Thus $y\ge M/(\epsilon-\delta)$ suffices when $\epsilon>\delta$. For $\epsilon<\delta$, averaging over one period of starting points rules out any uniform threshold. Equality also fails for real endpoints: take a marked integer $m$, $x=m-1/2$, and $y=qM+1$. The window contains $qC+1$ marked integers, exceeding $\delta(qM+1)$ for arbitrarily large $q$.

There is an exact finite reduction for integer half-open windows. Define

$$B(r)=\max_{u\bmod M}\sum_{j=1}^r b(u+j),\qquad 0\le r<M.$$

For integer length $L=qM+r$, the largest count is exactly $qC+B(r)$. A length is bad precisely when

$$B(r)-\epsilon r>q(\epsilon M-C).$$

If $\epsilon>\delta$, each remainder permits only finitely many bad $q$. For rational $\epsilon$, enumeration of these residues and inequalities therefore gives an exact finite algorithm for this endpoint convention. No enumeration has been run. For an integer open interval of length $L$, the count instead uses $L-1$ consecutive integers, while the comparison remains $\epsilon L$. In particular the equality case must be recomputed; the real-endpoint argument cannot be transferred unchanged.

The construction $M+d$, $d\in S$, gives $n-1$ consecutive marked integers. For $0<\epsilon\le1$, real windows containing this block and of lengths arbitrarily close from below to $(n-1)/\epsilon$ violate the density bound. Thus any real uniform threshold is at least $(n-1)/\epsilon$. Integer endpoints require rounding.

The resulting feasibility criterion does not determine sharp dependence on $n$ and $\epsilon$, especially when $\epsilon\downarrow\delta$. Ford's short-window theorem requires length at least $X/(\log z)^{10}$ in its specified range. This is not a threshold depending only on $n,\epsilon$ for arbitrary starting $X$. That theorem is used at statement depth; its proof and the separately reported linear-order claim are not audited here. The remaining problem is a sharp cyclic-discrepancy estimate with the endpoint convention fixed.